a ball with mass 0.15 kg is thrown upward with initial velocity 20 m/sec from the roof of a building 30 m high. there is a force due to air resistance of |v|/30, where velocity v is measured in m/sec.
a. find the maximum height above the ground the ball reaches.
b. find the time the ball hits the ground.
you cannot use the kinematic equations... you must use calculus. (which means you can't use average velocity!)
I answered this already by an approximate method. If you want to use differential equations, you are on your own. A numerical integration, using small time intervals and recalculating v, friction and incremental motion each time, would be much easier. A computer spreadsheet could be used.
To find the maximum height the ball reaches, we need to determine when the velocity of the ball becomes zero. We can then calculate the maximum height using the position function.
a. To find when the velocity becomes zero, we solve for the time (t) at which the velocity function equals zero. Let's start by determining the velocity function. The force due to air resistance can be modeled as F = -kv, where k is a constant and v is the velocity of the ball. If we consider the upward direction as positive, the net force acting on the ball can be written as:
ma = -mg - kv
Where m is the mass of the ball, g is the acceleration due to gravity, and a is the acceleration of the ball. Since the ball is moving upwards, a will be negative, and the equation becomes:
-ma = mg - kv
Rearranging the equation and substituting m and g with their respective values, we get:
ma + kv = mg
Now, we know that acceleration is the derivative of velocity with respect to time, so:
(dv/dt) + (k/m) * v = g
This is a first-order linear ordinary differential equation. To solve it, we separate the variables and integrate both sides:
∫(1/v) dv = ∫(g / (k/m)) dt
Integrating and simplifying, we get:
ln|v| = (g / (k/m))t + C
Where C is the constant of integration. Exponentiating both sides, we have:
|v| = e^((g / (k/m))t + C)
Now, in the given problem, the ball is thrown upward with an initial velocity of 20 m/s, so we can rewrite the equation as:
v = ±e^((g / (k/m))t + C)
To determine the value of C, we use the initial condition that v = 20 m/s when t = 0. Substituting these values into the equation, we get:
20 = ±e^((g / (k/m)) * 0 + C)
Since e^C is always positive, we can drop the ± sign. Thus, the equation becomes:
20 = e^C
Now, taking the natural logarithm of both sides:
ln(20) = ln(e^C)
ln(20) = C
So, the equation for the velocity becomes:
v = e^((g / (k/m))t + ln(20))
To find when the velocity becomes zero, we set v = 0 and solve for t:
0 = e^((g / (k/m))t + ln(20))
e^((g / (k/m))t) = 1/20
(g / (k/m))t = ln(1/20)
t = (1 / (g / (k/m))) * ln(1/20)
Next, we can find the maximum height by considering the position function. The position function is the integral of the velocity function with respect to time:
s(t) = ∫[v(t)] dt
To find the maximum height, we need to find when the velocity becomes zero. Let's substitute the values we found for v and t into the position function:
s(t) = ∫[e^((g / (k/m))t + ln(20))] dt
Integrating, we have:
s(t) = (m/k) * e^((g / (k/m))t + ln(20)) + C'
Using the initial condition that s = 30 m when t = 0, we can determine the value of C':
30 = (m/k) * e^((g / (k/m)) * 0 + ln(20)) + C'
30 = (m/k) * 20 + C'
C' = 30 - (m/k) * 20
Thus, the equation for the position becomes:
s(t) = (m/k) * e^((g / (k/m))t + ln(20)) + [30 - (m/k) * 20]
To find the maximum height, we substitute the time we calculated when the velocity becomes zero into the position function:
s(max) = (m/k) * e^((g / (k/m)) * [(1 / (g / (k/m))) * ln(1/20)]) + [30 - (m/k) * 20]
Simplifying and evaluating, we can find the maximum height above the ground the ball reaches.
b. To find the time the ball hits the ground, we need to find when the position of the ball becomes zero. We set s(t) = 0 and solve for t:
0 = (m/k) * e^((g / (k/m))t + ln(20)) + [30 - (m/k) * 20]
(m/k) * e^((g / (k/m))t + ln(20)) = (m/k) * 20 - 30
e^((g / (k/m))t + ln(20)) = (m/k) * 20 - 30 / (m/k)
(g / (k/m))t + ln(20) = ln((m/k) * 20 - 30 / (m/k))
t = [ln((m/k) * 20 - 30 / (m/k)) - ln(20)] / (g / (k/m))
Simplifying and evaluating, we can find the time at which the ball hits the ground.